Now, a negative t is outside our domain. The wavelength of this cycle is 4(pi) hours (not 2(pi) due to the "0.5" inside the cosine function), but cosine will return to the same value every half cycle. So add 2(pi) to this result:
t = 4.60105 hr. (I'll show you how to get this in terms of hours and minutes later.)

How to get the other solution? Well, whoever did the problem simply added another 2(pi) to get to the next cycle, but as you can see there is an earlier solution around 8 hours. Now, the cosine function is symmetric about its minimum point, here t = 2(pi) hrs (indicated by the dotted purple line). So there is a time difference of 2(*pi) hr - 4.60105 hr = 1.68214 hr between the first 25 ft depth and the minimum, so there will be another 25 ft depth at 2(pi) + 1.68214 hr = 7.96532 hr.